Question

How do you prove `sinx+cosx=(2sin^2x-1)/(sinx-cosx)`?

Answer 1

Try this:
`sinx+cosx=[2(1-cos^2x)-sin^2x-cos^2x]/(sinx-cosx)`
`(sinx+cosx)(sinx-cosx)=2-2cos^2x-sin^2x-cos^2x`
`sin^2x-cancel(cos^2x)=2-2cos^2x-sin^2x-cancel(cos^2x)`
`2sin^2x=2(1-cos^2x)`
`2sin^2x=2sin^2x`

Where I used the fact that:
`sin^2x+cos^2x=1`